Birth Death process An office has two employees that process incoming orders. these two are always busy and they process the orders at the rate of 100/day for each person. However they are smokers. On an average they will take a smoke break every 50 minutes (distributed exponentially) and it takes an average of 5 minutes to return to work (distributed exponentially). 
Calculate the Steady state probabilities that 0,1,2 are working.
Find the expected output of these people in a 25 day work period assuming an 8 hour work day.

I understand that this is a birth death process modelling machine breakdown. But I am not sure how to solve the same using the same.
 A: Let $\mu$ be the service rate, $\alpha$ the rate at which the workers take a break, and $\beta$ the rate at which the workers return from a break. Assume that the workers only take breaks when idle. Let $X(t)$ be the number of workers busy at time $t$, then $\{X(t):t\geqslant0\}$ is a continuous-time Markov chain on $\{0,1,2\}$ with transition rates
$$
q_{ij} = \begin{cases}
2\beta,& (i,j) = (0,1)\\
\beta,& (i,j) = (1,2)\\
\mu+\alpha,& (i,j) = (1,0)\\
2\mu,& (i,j) = (2,1).
\end{cases}
$$
We have the detailed balance equations
\begin{align}
2\beta\pi_0 &= (\alpha+\mu)\pi_1\\
\beta\pi_1 &= 2\mu\pi_2
\end{align}
which give
$$
\pi_1 = \frac{2\beta}{\alpha+\mu}\pi_0,\quad \pi_2 = \frac{\beta^2}{\mu(\alpha+\mu)}\pi_0,
$$
so from the condition $\pi_0+\pi_1+\pi_2=1$ we find that
\begin{align}
\pi_0 &= \frac{\mu(\alpha+\mu)}{\mu(\alpha+\mu)+\beta(\beta+2\mu)}\\
\pi_1 &= \frac{2\beta\mu}{\mu(\alpha+\mu)+\beta(\beta+2\mu)}\\
\pi_2 &= \frac{\beta^2}{\mu(\alpha+\mu)+\beta(\beta+2\mu)}.
\end{align}
Substituting the given rates (in units of $\mathrm{hour}^{-1}$) $\mu=25/2$, $\alpha=6/5$, and $\beta=20$ we have
$$\pi_0 = \frac{137}{857}, \quad \pi_1 = \frac{400}{857},\quad \pi_2 = \frac{320}{857} . $$
The expected output per day is  $\mu(\pi_1 + 2\pi_2)$, or $\frac{13000}{857}$. This gives an approximate expected output in a $25$-day work period of $379.2299$.
